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0

If you have $H = X\oplus Y$, then the projection onto $X$ is not well-defined without $Y$. The reason for this is that the splitting $H = X\oplus Y$ is not unique. That said, given $X$ there are many spaces $Y$ such that $H=X\oplus Y$. For $H=\mathbb R^2$, $X=span\pmatrix{1\\0}$, $Y_t=span\pmatrix{t \\1}$, we have $H=X\oplus Y_t$ for all $t\in\mathbb R$. ...


1

Since $g$ is clearly and odd function, and cosine is even and sine is odd, the coefficient of cosine will have to be $0$, and that's what you got. And you can check by subsitution whether your two linear equations are satisfied by the answer you got.


0

The reason you want to specify "along $Y$" is that you may have to deal with multiple orthogonal projections at the same time, and then it is useful to have a way to speak of different projections without too much fuss. For example, this occurs in the Gram-Schmidt orthonormalization process, where you have to iteratively take a projection along a ...


1

Obviously, the norm is at least what you anticipate by setting $f = e_n$ where $n$ almost realizes the supremum of $|\lambda_n - \lambda|^{-1}$. Conversely, write $f$ as $f = \sum_n c_n e_n$. Then $$\|(K - \lambda I)^{-1} f\|^2 = \sum_n |c_n|^2 |\lambda_n - \lambda|^{-2} \leq (\sup_n |\lambda_n - \lambda|^{-2}) (\sum_n |c_n|^2).$$ Taking the square root ...


1

When dealing with functionals, I prefer to write the derivative as $F'(x;h)$, meaning $$F(x+h)=F(x)+F'(x;h)+o(\|h\|)\tag{1}$$ So, $F'(x;\cdot )$ is a linear functional on $\ell^2$. To find it formally (without justification), just differentiate $F$ coordinatewise: $$F(x;h) = \sum_{n=1}^\infty\frac{2x_n }{n^3} h_n-\sum_{n=1}^\infty 4x_n^3 h_n \tag{2}$$ Then ...


1

You don't need to show $x\in \ell^2$; that is a part of the set-up. The equality $y=Tx$ can be checked component-wise: you need to check that $y_n=(Tx)_n$ for every $n$. That is, $y_n=nx_n$. Here are the two steps leading to the goal: $$x_n =\lim_{m\to\infty} x^{(m)}_n\tag{1}$$ $$y_n =\lim_{m\to\infty} (Tx^{(m)})_n\tag{2}$$


-1

Ok, I think I got it now... Both work perfectly fine as they are always nondegenerate: $$\mathcal{A}_\text{CAR}:\quad a(f\neq0)\neq0$$ $$\mathcal{W}:\quad W(f)\neq0$$ A counterexample is provided by the angular momentum algebra: $$\mathcal{J}:\quad [J_i,J_j]=\imath\varepsilon_{ijk}J_k$$ There one has one trivial representation: $J_x=J_y=J_z=0$


0

Ok, I think I finally got it... (However, if anybody finds bugs, typos, loopholes etc. then please let me know. Thanks!) Framework Given the natural numbers $\Omega:=\mathbb{N}$ and the Hilbert space $\mathcal{H}:=\ell^2(\mathbb{N})$. Choose the canonical basis by: $e_n:=\chi_n$ Consider the spectral measure $E(\{n\}):=P_n$. Operator Domain Regard the ...


3

Hint Note that $$<T(x_1+x_2),y>=<x_1+x_2,Ty>=<x_1,Ty>+<x_2,Ty>\\=<Tx_1,y>+<Tx_2,y>=<Tx_1+Tx_2,y>, \forall y \in \mathbb{H},$$ from where $T(x_1+x_2)=T(x_1)+T(x_2).$ Proceed in the same way with product by scalars.


0

Define $A:H^s_0 \to (H^s_0)^*$ by $\langle Au, v \rangle = (u,v)_{H^s_0}$ and consider the equation $\langle Au, v \rangle = (f,v){L^2}.$ $u$ exists uniquely, and the solution map $T(f) = u$, $T:L^2 \to L^2$ is compact. Then one can apply the Hilbert-Schmidt theorem.


0

That's a comment but it got a bit too long. As far as I know, classically, the approximation property deals with bounded linear operators. But let us assume your point for a minute. If it is not "bounded linear" then what is you definition of "bounded" operator? E.g., 1) a bounded function, that is a function with bounded range. But a bounded linear ...


2

$x\mapsto\langle x, h\rangle$ is a continous linear functional on $H_0$, therefore, if $H_0, \langle, \rangle_0$ is a Hilbert space, this follows from a well known theorem of Riesz. (See http://en.wikipedia.org/wiki/Riesz_representation_theorem)


0

I seem to have found the solution. Correct me if I'm wrong, but if $ h_1, h_2 \in H_0 $ we can say that: $$\langle Th_1, h_2 \rangle = \langle h_1, T^*h_2 \rangle = \langle Ph_1, T^*h_2 \rangle = \langle h_1, PF^*h_2 \rangle$$


1

An operator $P$ on a Hilbert space is an orthogonal projection iff $$ P^{2}=P=P^{\star}. $$ Such a projection always has a closed range because $\mathcal{R}(P)=\mathcal{N}(I-P)$. You can try to define the orthogonal projection $P_{\mathcal{M}}x$ of a vector $x$ onto a subspace $\mathcal{M}$ as the unique $m \in \mathcal{M}$ such that ...


0

Take two lineraly independent discontinuous linear functionals $f,f'$. The intersection $I=H\cap H'$ of their kernels is dense of codimension $2$. If you take a vector $X$ that is in the kernel $H$ of the first linear functional and not in $H'$, then $$I\oplus \Bbb R X=H$$ and both $I$ and $H$ are dense.


0

Aah, sometimes things are so simple. :) Suppose it vanishes: $a(f_0)=0$ Then one has by the CAR relations: $$0=\{a(f),a(f)^*\}=\|f\|^2\neq0$$ That is a contradiction!


2

For a concrete example of an operator that isn't even closable, consider $$ M: (x_1,x_2,x_3,\ldots,x_n,\ldots)\mapsto(x_1,2x_2,3x_2,\ldots,nx_n,\ldots)$$ defined on the subspace of sequences with bounded support in $\ell^2$. Then the sequence $(\frac 1n \mathbf e_n)_n$ clearly converges (to 0), but $(M(\frac 1n\mathbf e_n))_n = (\mathbf e_n)_n$ doesn't.


1

Any bounded operator $A$ on a dense vector subspace $D$ of a Hilbert space $H$ such that $D\neq H.$


1

A basic neighbourhood of a point $p$ in the weak topology can be written in the form $$U(y_1, \ldots, y_n; p) = \{x: \left|\langle y_j, x - p \rangle\right| < 1, j = 1 \ldots n\}$$ where $\{y_1, \ldots, y_n\}$ is any finite set of vectors in $\mathcal H$. If $\|p \| < 1$, find $v$ so $\langle y_j, v\rangle = 0$ for $j = 1\ldots n$ and add an ...


2

Let's take the simplest possible system: a one-dimensional spinless particle. We can describe it in terms of wave functions $\psi\in L^2(\mathbb{R})$. The fundamental observable is "position" $x$, which must be represented as self-adjoint operator acting on wave functions. Quantum mechanics tells us that the right way of doing so is multiplying by $x$. ...


3

Not true. Any linear operator defined on a linear subspace can be extended to the whole space, but You need some form of the Axiom of Choice to do this. You lose any nice properties of the operator (e.g. if the operator is symmetric, the extension won't be).


0

Let $X=C[0,1]$ with the uniform metric, $Y=C^1[0,1]$ and $D\colon Y\to X$ the derivative operator: $Df=f'$ for all $f\in Y$. $Y$ is dense in $X$ and $D$ cannot be extended to $X$ as a continuous operator.


2

A Banach space is a complete metric space with a norm, while a Hilbert space is a complete metric space with an inner product. An inner product induces a norm by $||x|| = \sqrt{<x,x>}$, which means that any Hilbert space is also a Banach space. However, the converse need not hold as the norm isn't always expressable in terms of the inner product. An ...


3

The adjoint is defined as $$ \langle T^\ast u, v\rangle = \langle u, Tv\rangle. $$ This only defines $T^\ast u$ uniquely if the set of all $v$, for which this equation makes sense is dense. But this means that $T$ has to be densely defined.


1

It is only possible with infinite dimensions. A good example of a Hilbert space is a set of real or complex sequences for which the sum of squares absolutely converges. The scalar product of two sequences $x=(x_n)$ and $y=(y_n)$ is $$\sum_{n=0}^{\infty}x_n\overline{y_n}$$ Let's call $x$ finite if there is such $n$ that for each $i>n$ the following ...


4

Let $H = L^2([0,1])$ and let $P$ be the set of polynomials in $H$.


6

Consider the space of square summable sequences and the subspace of sequences whose only finitely many terms are different from zero. This subspace is dense and is not the whole space hence it cannot be closed.


1

on p.475 Kolmogorov-Fomin says: "$\dots0$ is the only possible accumulation point for the sequence $\{\mu_n\}\dots$" Kolmogorov-Fomin does not exclude the possibility that $\sigma(A)$ is finite and there is no accumulation point for $\sigma(A).$


1

This long answer is based on many little observations; at every step I will use the preceding ones, sometimes implicitly. Tell me if some step is not clear. Call $P_1:=P_{C_1}$ and $P_2:=P_{C_2}$. You can assume that $C_1$ is compact (if the compact set is $C_2$, with the following argument you will get that $P_2 P^n(x)$ converges to some $\ell$ fixed by ...


1

One possibility is the following: Decompose $K =K_+ - K_-$ and $\phi =\phi_+ -\phi_-$, where $K_\pm$ are the positive and negative parts of $K$. Then your integrand is $$ (K_+ - K_-)\cdot (\phi_+-\phi_-)= K_+ \phi_+ + K_- \phi_- -(K_+ \phi_- + K_- \phi_+). $$ Now each of the contents of the two brackets is a measurable, nonnegative function. Fubini's ...


1

The argument can be made rather simpler: if you have a given ON basis $e_i$, $i\in I$, the vector $v=\sum_{n\in {\bf N}} 2^{-n}e_{i_n}$, where $i_n$ are distinct, is clearly not a finite linear combination of the $e_i$ (because if we subtract from $v$ some finite linear combination of $e_i$, we can still find some $i_n$ such that the result is not orthogonal ...


2

The thing you're missing (I think - looks like you're almost there in your edit) is Schwarz's inequality. We have that \begin{equation} |A\phi_1 - A\phi_2|_{L_2} = \lambda^2\int_a^b \left|\int_a^s K(s,t)(\phi_1(t)-\phi_2(t))\mathrm{d}\mu_t\right|^2\mathrm{d}\mu_s. \end{equation} Then by Schwarz's inequality we can bound the inner integral squared as the ...


1

Have you tried using Cauchy-Schwartz? $$|\langle Tx, y \rangle | \ \leq \ ||Tx|| \, || y || \ \leq \ ||Tx||$$ Is there a $y$ for which this inequality is equality?


2

Since the $H^k$ are descending, having $H^m = H^n$ for some $n > m$ means $H^k = H^m$ for all $k \geqslant m$, as $H^m \supset H^{m+1} \supset \dotsc \supset H^n = H^m$ forces all these spaces to be equal, and then, for $k \geqslant m$ we have $H^{k+2} = T(H^{k+1}) = T(H^k) = H^{k+1}$ by induction. Suppose that $H^1\subsetneq H$, but the $H^k$ are not ...


2

Since $\ker T$ and $\operatorname{im} T^\ast$ are closed and orthogonal to each other, the subspace $$E := \ker T \oplus \operatorname{im} T^\ast$$ of $H$ is closed. Thus we have $E = H$ if an only if $E$ is dense. But a subspace $F$ of $H$ is dense, if and only if $F^\perp = \{0\}$. So showing that $$(\ker T \oplus \operatorname{im} T^\ast)^\perp = (\ker ...


1

Let $(\phi_i)_i$ be an orthonormal basis of $L^2(a,b)$. Define the basis $(\Phi_{ij})_{i,j}$ of $L^2([a,b] \times [a,b])$ via tensor products, $\Phi_{ij}(s,t) = \phi_i(s) \phi_j(t)$. Now expand the kernel $K$ in this basis, $$ K = \sum_{i,j} \alpha_{ij} \Phi_{ij}. $$ Truncate the expansion and you have a degenerate approximation: $$ K_N(s,t) = \sum_{i,j \le ...


-1

Put $a_n=\frac{1}{n}$. Then $ \sum \frac{1}{n^2}=\frac{\pi^2}{6}$,finite. But $\sum n^2$ is not in $l^2$.


6

If $a\in\ell^2$, then $a_n\to0$. Thus $1/a_n\not\to0$, which prevents it from being in $\ell^2$.


0

Note that $$\left(\sum_{\nu =1}^n 1\right)^2=n^2\leqslant \sum_{\nu=1}^n a_\nu^2\sum_{\nu =1}^n a_\nu^{-2}$$ This gives a pretty rough estimate of how the sum of the squares of the reciprocals diverge. Of course the simplest is Martin's approach.


1

It sounds like you are focused on the closedness of $H$ rather than the subspace part, so I'll address just the closedness. (Note that if you've already established $H$ is a subspace, then $H$ is necessarily a vector space itself.) Let $\{x_n\}$ be a Cauchy sequence in $H$ (Cauchy with respect to whatever norm you have on $H$). Then $x_n\to x$ for some $x$. ...


2

It's not that the domain $\mathcal{D}(A)$ of a closed operator $A$ is closed in $H$, but that the graph $$\mathcal{G}(D) := \{(\xi,A\xi) \in \mathcal{H} \oplus \mathcal{H} \mid \xi \in \mathcal{D}(A)\}$$ is closed in $\mathcal{H} \oplus \mathcal{H}$. Indeed, your second statement, which is just the definition of a closed operator, actually says that if ...


2

(1) is easier to asnwer: If $X$ is a metric space and $Y \subset X$ is a dense subset, then $\overline Y = X$. So a proper closed subspace cannot contain a dense subspace. (2) is just a definition, it does not say that $D(A)$ is closed. It says that $A$ is called closed if and only if the graph $\{(x, Ax): x\in D(A)\}$ is closed. $D(A)$ is Not closed. ...


1

For your first question, no $H_0^1(\mathbb{R}^n)$ is not a Hilbert space with the inner product $(u,v)=\int \nabla u \cdot \nabla v$. Its completion however is, and is usually denoted by $\mathcal{D}^{1,2}(\mathbb{R}^n)$ or $L^{1,2}(\mathbb{R}^n)$ and consists of functions in $L^{2^*}(\mathbb{R}^n)$ with integrable gradient, where $2^*=2n/(n-2)$ is the usual ...


1

You've shown that $\langle (T-I)y,x\rangle=0$ for all $y\in D(T)$. Another way to say that is that $\langle z,x\rangle=0$ for all $z\in R(T-I)$. Now use the fact that $R(T-I)$ is dense in $H$ to conclude that $\langle z,x\rangle=0$ for all $z\in H$, which is what you need to say that $x=0$.


1

Here is what I thought. I am not very sure but I am happy to discuss with you. The way we cast $\Delta^2u=0$ into a weak formulation, so that we could use Lax-Milgram, tells us that $H_0^2$ is a suitable space. Suppose we have a nice solution already, then we test $\Delta^2u=0$ with a $C^\infty$ function $v$ and see what happens. We have $$\int_\Omega ...


0

The above calculations apply as well for contractions: $$z\in\mathcal{H}\otimes\mathcal{H}:\quad\|(S\otimes T)z\|^2=\|(S\otimes T)\sum_{i=1}^nx_i\otimes y_i\|^2=\|\sum_{i=1}^n(Sx_i)\otimes(Ty_i)\|^2\\\leq\sum_{i=1}^n\|(Sx_i)\otimes(Ty_i)\|^2=\sum_{i=1}^n\|Sx_i\|^2\|Ty_i\|^2\leq\sum_{i=1}^n1\cdot\|x_i\|^21\cdot\|y_i\|^2=\|\sum_{i=1}^nx_i\otimes ...


1

For #1: Let's call your isomorphism $T$, the isometry from $\mathcal{H}_K$ to $L^2(P)$ which maps $K_t$ to $B_t$. Yes, $T$ really is a stochastic integral: $Tf$ is the Itô integral, not of $f$, but of $f'$. So $$Tf = \int_0^1 f'(t)\,dB_t$$ or in other words, for deterministic $g \in L^2([0,1])$, $$\int_0^1 g(t)\,dB_t = T\left(\int_0^\cdot ...


1

That element you want to form is just an elementary tensor $x\otimes y$ in the algebraic tensor product $\mathcal H\otimes\mathcal H$. Then you want to have sums of those guys, and then limits of them.


0

I realized that $u=0$ on $\partial\Omega$ and $\nabla u\cdot \nu=0$ imply $\nabla u=0$ on $\partial\Omega$ for smooth $u$. So taking $H^2_0$ as space where we seek the weak solution, i.e. intuitively requiring $\nabla u=0$ on $\partial\Omega$, is NOT more than $\nabla u\cdot \nu =0$


0

Bochner Since it is separable valued: $$\varphi\in\mathcal{C}(\mathbb{R},E):\quad\mathbb{R}\text{ separable}\implies (\alpha\varphi)(\mathbb{R})\text{ separable}$$ and weakly measurable: $$l\in E':\quad(\alpha\varphi)\text{ continuous}\implies l\circ(\alpha\varphi)\text{ measurable}$$ so by Pettis' criterion strongly measurable: $$\varphi\text{ Bochner ...



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